Question: $\overline{AB}$ = $\sqrt{145}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{145}$ $?$ $ \sin( \angle BAC ) = \frac{8\sqrt{145} }{145}, \cos( \angle BAC ) = \frac{9\sqrt{145} }{145}, \tan( \angle BAC ) = \dfrac{8}{9}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{145}} $ $ \overline{BC}=\sqrt{145} \cdot \sin( \angle BAC ) = \sqrt{145} \cdot \frac{8\sqrt{145} }{145} = 8$